Question: Solve for $x$ : $ 6|x - 2| - 4 = 1|x - 2| + 2 $
Explanation: Subtract $ {1|x - 2|} $ from both sides: $ \begin{eqnarray} 6|x - 2| - 4 &=& 1|x - 2| + 2 \\ \\ { - 1|x - 2|} && { - 1|x - 2|} \\ \\ 5|x - 2| - 4 &=& 2 \end{eqnarray} $ Add ${4}$ to both sides: $ \begin{eqnarray} 5|x - 2| - 4 &=& 2 \\ \\ { + 4} &=& { + 4} \\ \\ 5|x - 2| &=& 6 \end{eqnarray} $ Divide both sides by ${5}$ $ \dfrac{5|x - 2|} {{5}} = \dfrac{6} {{5}} $ Simplify: $ |x - 2| = \dfrac{6}{5}$ Because the absolute value of an expression is its distance from zero, it has two solutions, one negative and one positive: $ x - 2 = -\dfrac{6}{5} $ or $ x - 2 = \dfrac{6}{5} $ Solve for the solution where $x - 2$ is negative: $ x - 2 = -\dfrac{6}{5} $ Add ${2}$ to both sides: $ \begin{eqnarray} x - 2 &=& -\dfrac{6}{5} \\ \\ {+ 2} && {+ 2} \\ \\ x &=& -\dfrac{6}{5} + 2 \end{eqnarray} $ Change the ${ + 2}$ to an equivalent fraction with a denominator of $5$ $ x = - \dfrac{6}{5} {+ \dfrac{10}{5}} $ $ x = \dfrac{4}{5} $ Then calculate the solution where $x - 2$ is positive: $ x - 2 = \dfrac{6}{5} $ Add ${2}$ to both sides: $ \begin{eqnarray} x - 2 &=& \dfrac{6}{5} \\ \\ {+ 2} && {+ 2} \\ \\ x &=& \dfrac{6}{5} + 2 \end{eqnarray} $ Change the ${ + 2}$ to an equivalent fraction with a denominator of $5$ $ x = \dfrac{6}{5} {+ \dfrac{10}{5}} $ $ x = \dfrac{16}{5} $ Thus, the correct answer is $x = \dfrac{4}{5} $ or $x = \dfrac{16}{5} $.